Solve the problem.
-A coil of wire rotating in a magnetic field induces a voltage given by $\mathrm { e } = 20 \sin \left( \frac { \pi \mathrm { t } } { 4 } - \frac { \pi } { 2 } \right)$
where $t$ is time in seconds. Find the smallest positive time to produce a voltage of $10 \sqrt { 2 }$.
A) $3 \mathrm { sec }$
B) $2.8 \pi \mathrm { sec }$
C) $3 \pi \mathrm { sec }$
D) $2.8 \mathrm { sec }$

Question

Quizplus · Accepted Answer

To find the smallest positive time $t$ that produces a voltage of $10 \sqrt{2}$, we set the given equation equal to $10 \sqrt{2}$:$$20 \sin \left( \frac{\pi t}{4} - \frac{\pi}{2} \right) = 10 \sqrt{2}$$Dividing both sides by 20:$$\sin \left( \frac{\pi t}{4} - \frac{\pi}{2} \right) = \frac{10 \sqrt{2}}{20} = \frac{\sqrt{2}}{2}$$The sine function equals $\frac{\sqrt{2}}{2}$ at angles of $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ within one cycle of sine. However, due to the phase shift of $-\frac{\pi}{2}$, we adjust the equation to find $t$:$$\frac{\pi t}{4} - \frac{\pi}{2} = \frac{\pi}{4}$$Solving for $t$:$$\frac{\pi t}{4} = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$$$$t = 3$$

Solve the Problem e=20sin⁡(πt4−π2)\mathrm { e } = 20 \sin \left( \frac { \pi \mathrm { t } } { 4 } - \frac { \pi } { 2 } \right)e=20sin(4πt​−2π​)

Solve the Problem $\mathrm { e } = 20 \sin \left( \frac { \pi \mathrm { t } } { 4 } - \frac { \pi } { 2 } \right)$